Question

Решите систему уравнений:

3x+y+4=0

X^2-y^2=2

Answer 1

{3x+y+4=0
{x²-y²=2

{y=-3x-4
{x²-y⅜=2

x²-(-3x-4)²=2

x=-3/2

y=-3*(-3/2)-4

y=1/2

(x.y)=(-3/2,1/2)

{3*(-3/2)+1/2+4=0
{(-3/2)²-(1/2)²=2

{0=0 => }2=2

(x.y)=(-3/2,1/2)